Question

Medal help...I'm confused...
A deck of cards consists of 8 blue cards and 5 white cards. A simple random sample (random draws without replacement) of 6 cards is selected. What is the chance that one of the colors appears twice as many times as the other?

Answer 1

List the ways in which you could have (1) twice as many blue cards as white cards (e.g. 4 blue 2 white) and (2) twice as many white cards as blue cards (e.g. 4 white 2 blue).

Then calculate the probability of each happening. Remember they're done without replacement, so after drawing the first card, there are only 12 cards left etc. And the order doesn't matter. So for 4 blue 2 white it's basically
((8*7*6*5)*(4*3))/(13*12*11*10*9*8).

Or you can just find the number of ways it can happen and divide it by the total no. of possible outcomes. (13!)

Answer 2

ok i'll try that...thanks

Answer 3

w8...this is what i did before...13/13*7/12*6/11*5/10*5/9*4/8 that's 4 blue and 2 whites
13/13*4/12*3/11*2/10*8/9*7/8...that's or 4 whites and 2 blues

Answer 4

and i got the answers for both...do i have to add it or
multiply

Answer 5

coz i got .0442 on 1 combination and .01414 on the other

Answer 6

you are putting an order on the blue and white cards by doing it the way you are

Answer 7

what do you mean by that...i don't get it

Answer 8

for the 4 blue and 2 white ...

are you looking to get the first 4 blue and the second 2 white or do you just want 4 blue and 2 white?

Answer 9

just 4 blue and 2 whites

Answer 10

then you need to modify how you are computing the answer

Answer 11

thank you

Answer 12

I would use \({n\choose r}\) to help compute the answer

Answer 13

or \(_nC_r\) if you are not familiar with the above notation

Answer 14

so if you just wanted 4 blue and 2 white you would do this

\[\frac{{8\choose 4}{5\choose 2}}{{13\choose 6}}\]

Answer 15

I'm having problem with prob calculator ( I dont know why maybe because o the laguague barrier..i really dont know) that's why I'm doing most of the problem the long way

Answer 16

you can do it the long way but you have to be mindful of accidentally putting an order on the section process...it is very easy to mess up.

Answer 17

if you have \[\frac{(8\times7\times6\times5)\times(5\times4)}{13\times12\times11\times10\times9\times8}\]

then you have put an order on the cards

Answer 18

ie you just fond the prob of getting in order b,b,b,b,w,w

Answer 19

there are \[\frac{6!}{4!2!}\] ways to permute the b's and w's

Answer 20

so your answer (atlest for 4 blue and 2 white ) is

\[\frac{(8\times7\times6\times5)\times(5\times4)}{13\times12\times11\times10\times9\times8}\times\frac{6!}{4!2!}\]

Answer 21

using shorthand...that is

\[\frac{{8\choose 4}{5\choose 2}}{{13\choose 6}}\]

Answer 22

do i need to get the prob of both combination or just choose 1 combination

Answer 23

you need to calculate

P(4B and 2W)+P(2B and 4W)

if that is what you are asking

Answer 24

ok..got it..thanks a lot

Answer 25

Zarkon...can you tell me if I did this right this time...

I roll a die repeatedly. Find the chance that the first 4 rolls all show different faces, and the 5th roll shows a face that has appeared before.

6/6*5/6*4/6*3/6*4/6